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ASTRONOMY: 



COMPILED FROM THE LECTURES 



PROFESSOR HACKLEY, 



COLUMBIA COLLEGE. 



PUBLISHED BY THE JUNIOR CLASS— 1855 




NEW YORK: 
BAKER, GODWIN & CO., BOOK AND JOB PRINTERS, 

CORNER NASSAU AND SFRUCE STREETS. 

1855. 







Entered according to Act of Congress, in the year 1855, by Charles N. 
Clark, in the Clerk's Office of the District Court of the United States for the 
Southern District of New York. 






:••" 



V 



ASTRONOMY. 



Part First. Questions on Astronomy. 

1. Into what classes is Astronomy divided ? 

(1.) Practical Astronomy, which includes the use of in- 
struments, and the application of the mathematical princi- 
ples, necessary for the determination of the places of the 
heavenly bodies, or the co-ordinates of their positions, to- 
gether with the necessary corrections of these co-ordinates. 

(2.) Theoretic Astronomy, or the theory of the motions 
of the heavenly bodies, and the nature and position of their 
orbits, derived from the results of practical astronomy, by 
the aid of the theory of conic sections. 

(3.) Descriptive Astronomy, which comprehends the 
physical constitution and phenomena presented by the heav- 
enly bodies, individually. This branch requires the aid of 
a powerful telescope : the most distinguished names that 
occur in its history and development are the two Herschels, 
and Schroeder, of Lilienthal. 

(4.) Physical Astronomy, which seeks to investigate all 
the movements of the material universe, as a purely mechan- 
ical problem. It is also known under the name of " Celes- 
tial Mechanics," and has been illustrated by La Place and 
La Grange. This branch involves the highest mathematics. 

2. What are the proofs of the figure of the Earth f 
(1.) It has been circumnavigated. 

(2.) As a traveler proceeds directly north or south, the 
same stars cross the meridian higher or lower in the diurnal 
1 



2 ASTRONOMY. 

motion, the change being gradual as the traveler advances ; 
this indicates a gradual curvature of the earth in this direc- 
tion. 

(3.) The circular form of the horizon at sea. If the 
earth were a sphere, and the visual ray drawn from the eye 
of the observer, tangent to the surface of the earth, were set 
in rotation about the diameter of the earth, passing through 
the eye as an axis, it (the visual ray) would describe a visual 
cone, whose circle of contact with the sphere would be the 
boundary of the visual horizon. The identity of the figure 
with that actually observed is a proof of the spheroidal fig- 
ure of the earth. 

(4.) The form of the earth shown in eclipses of the 
moon. 

(5.) The circumstances attending the arrival of ships 
from sea. At first, only the tops of the masts are observed, 
then we see the yards, one after another, until, at last, as 
they come nearer, we discern the whole fabric. 

(6.) The analogy of form, between all the other bodies, 
composing the solar system. 

(7.) The exact spheroidal figure of the earth is ascer- 
tained by minutely accurate measurements in the direction 
of meridians and parallels, together with astronomical obser- 
vations. See Dr. HacTdefs Trigonometry, p. 366. 

3. What are the proof s of the EartKs Rotation? 

(1.) The apparent diurnal rotation of the sun, moon, 
planets, and fixed stars, at such immensely varying distances 
from the earth, make the probabilities a thousand to one 
that it is the earth that is in rotation on its axis, rather than 
that such an immense number of objects, so vast and so re- 
mote, should have their motions of translations so exactly 
adjusted in proportion to their distance from the earth, as 
to conspire in producing the effects we see. 

(2.) When a body is dropped from the top of a high 
tower, it will not fall exactly at the base of the tower, or in 
a perpendicular to the surface of the earth from the point 
whence it was allowed to fall, but it will fall a little east of 
the perpendicular, the velocity communicated at that greater 
height carrying the body further on, than if it had remained 
at the surface of the earth. 



MISCELLANEOUS QUESTIONS. O 

(3.) The pendulum experiment. By the property of in- 
ertia, the tendency of matter is to continue a motion, which 
it has received by the application of some force, in a con- 
stant direction, unless some extraneous force change it. 
Thus, a pendulum, set in motion in a certain plane, would 
ever oscillate in that plane, if the point of suspension be per- 
fectly free to revolve on a pivot; or if, instead of a rod, a 
thread without tortion be used, its pendulum will not feel 
the rotation of the point of suspension, or of the surface of 
the earth under it, with which it is connected in the diurnal 
motion. Now, if a circle be described about the horizontal 
projection of the point of suspension, in a manner like the 
dial of a clock, the points of the dial over which the pendu- 
lum oscillates will be found to slowly change, and to com- 
plete the whole circumference (the amount depends on the 
latitude) in about twenty-four hours. 

4. How is an orbit given ? 
By its elements. 

5. How many elements of an orbit are there? 
Seven. 

6. Which tivo determine the position of the plane of the 
orbit in space ? 

The longitude of the node and the inclination. 

7. Which two elements determine the size and shape of an 
orbit in its own plane? 

The semi-major axis (called also the mean distance) and 
the eccentricity. 

8. What element determines the position of an orbit in 
space f 

The longitude of the perihelion or vertex of the ellipse 
nearest the focus in which the sun is placed. 

9. JVhat tivo determine the position of the planet in its 
orbit at any given time ? 

The epoch of the perihelion passage and the periodic 
time. 

10. Which tivo of the seven elements do not occur in the 
orbit of the earth ? 

The longitude of the node and the inclination. 

11. What is the node of a plane f s orbit? 



4 ASTRONOMY. 

It is tlie point in which the orbit of the planet intersects 
the orbit of the earth, i. e., the plane of the ecliptic. 

12. What is that node called, through which a planet 
passes in going from the south to the north side of the 
ecliptic ? 

The ascending node.( Q ) 

13. What is that called, passed through in the passage 
from north to south? 

The descending node ( £5 •) 

14. What is the line of nodes? 

It is a line joining the nodes of a planet's orbit, and 
which always passes through the sun, or it is the intersec- 
tion of the plane of the planet's orbit with the plane of the 
ecliptic. 

15. What is the longitude of the node? 

The angle which the line of nodes makes with the line 
of equinoxes. (This angle being given, and a point on the 
plane of the ecliptic, viz.: the position of the sun, through 
which it passes, the position of this line will be entirely de- 
termined ; and then the angle, which the plane of the orbit 
passing through this line makes with the plane of the eclip- 
tic, entirely determines the position of the plane of the orbit 
in space.) 

16. What is the line of equinoxes? 

It is a line drawn from the sun to the equinoctial point. 
(A line drawn from the earth would be sensibly parallel to 
this, since the equinoctial point may be considered as at an 
infinite distance in comparison with the distance of the earth 
from the sun. The latter line is also the line of intersection 
between the plane of the equator and the plane of the 
ecliptic.) 

17. What is the longitude of the perihelion ? 

It is the angle which the projection of the major axis of 
the orbit, on the plane of the ecliptic, makes with the line 
of equinoxes. (This angle being given, the position of this 
projection, since it must pass through the center of the sun, 
is known on the plane of the ecliptic. A plane, through 
this projection, perpendicular to the plane of the ecliptic, 
will intersect the plane of the orbit, given in position by the 
first two elements as the major axis of the orbit ; the posi- 



MISCELLANEOUS QUESTIONS. 5 

tion of which, and therefore the orbit in its own plane, thus 
becomes known.) 

18. What is the epoch of the perihelion passage ? 

It is the instant at which the planet passes the perihe- 
lion of its orbit. (The perihelion is the point of the orbit 
nearest the sun, or to that vertex of the ellipse next to that 
focus on which the sun is placed. 

19. What is the radius vector of a planet? 

It is the distance from' the center of the sun to the point 
in the orbit which the planet has reached. 

20. What is the "anomaly?" 

It is the angle which the radius vector makes with the 
major axis ; or, it is the angular distance of the planet from 
the perihelion. This is known as the " true anomaly." 

21. What is the " mean anomaly ?" 

It is the angular distance from the perihelion which a 
planet would have, at any given time, if it moved with an 
uniform, velocity, and completed its period in the same time 
that it does with a varying velocity. 

22. To what is the true angular velocity of a planet in 
its orbit proportional '? 

It is inversely as the square of the radius vector ; i. e.: 
the square of the distance from the sun. It is, therefore, 
greatest when the planet is at the perihelion, and least when 
the planet is at the aphelion, or other vertex of the ellipse. 

23. How, from the sixth and seventh elements in a 
plane? s orbit, is the mean place of the planet found ? 

The " mean anomaly" will be that portion of three hun- 
dred and sixty degrees which the time elapsed, at the in- 
stant for which the mean position is required, since the 
epoch of the perihelion passage, is of the whole periodic 
time. 

24. How is the true place of the planet in its orbit, or 
the " true anomaly" found from the " mean ?" 

By Kepler's problem, which gives the means of de- 
termining a third " anomaly," called the " eccentric," from 
the " mean," and the means of determining the " true" from 
this " eccentric." 

25. What is the heliocentric latitude of a heavenly 
body ? 



6 ASTRONOMY. 

It is the angle which a line drawn from the center of 
the sun to the body, (the radius vector in case of a planet,) 
makes with the plane of the ecliptic. 

26. What is the grocentric latitude of a heavenly body? 
It is the angle which a line drawn from the center of 

the earth to the body makes with the ecliptic. 

27. What is the heliocentric longitude of a heavenly 
body ? 

It is the angle which the projection of the line drawn 
from the center of the sun to the heavenly body, (on the 
plane of the ecliptic,) makes with the line drawn from the 
center of the sun to the vernal equinox. 

28. What is the grocentric longitude ? 
It is the angle made by the projection of a line drawn 

from the center of the earth to the heavenly body, (on 
the plane of the ecliptic,) with the line of equinoxes. 

29. What is the apparent diameter of a heavenly body ? 
It is the angle subtended at the eye of the observer by 

the diameter of the disc of the body. 

30. What is the line of apsides? 
The line joining the perihelion and aphelion of an orbit, 

or it is the major axis of the orbit. 

31. What are the apsides of an orbit ? 
The perihelion and aphelion. 

32. Wkmt is called the lower apsis ? 
The perihelion. 

33. What the upper apsis ? 
The aphelion. 

34. What is the equation of the center ? 

It is the difference between the "true" and "mean" 
anomalies, i. e., the d4fe£e«€e between the true and mean 
radii vectores. 

35. How is the " mean anomaly" obtained from tables 
of the sun ? 

By taking the difference between the mean longitude of 
the sun and the mean longitude of the perigee. 

36. What is a sidereal year ? 

The interval between the instants at which the sun has 
the same difference of right-ascension from any star. 



<^ 






MISCELLANEOUS QUESTIONS. 7 

37. How long is this interval ? 
At the mean, 365.256 336 days. 

38. How does this compare with the tropical year ? 
It is somewhat longer. 

39. To what is this owing ? 
The precession of the equinoxes. 

40. What is the physical cause of precession ? 

The spheroidal figure of the earth, (produced by a pro- 
tuberant mass at the equator,) acted upon by the combined 
attraction of the sun and moon. 

41. What is the effect of this attraction? 

To produce a slow revolution of the axis of the equator 
about the axis of the ecliptic. 

42. What is the expression for the period of revolution ? 

360° 

jttttt. = m round numbers, 26,000 years. 

50. 2 J 

43. What produces the precession of the equinoxes? 
The revolution of the axis of the equator, carrying the 

equator with it, and retaining its plane always perpendicular. 

44. Into how many classes are the disturbances of ele- 
ments divided ? 

Two. 

45. What are they ? 
Secular and periodic. 

46. What difference is there betiveen them ? 

The periods in which the secular complete themselves 
are very long ; those of the periodic much shorter. 

47. What must be done, in order to have the exact value 
of any element, as the obliquity of the ecliptic, for instance, 
at any given time ? 

The combined effect of each of these two classes must 
be computed. 

48. Where would you find the mean obliquity, obtained 
in this way, at the beginning of the year ? 

In the Nautical Almanac, just before the ephemeres of 
the planets. 

49. In what does the actual obliquity differ from the 
mean ? 

By the effect of the mutation. 



ASTRONOMY. 



Part Second. Problems. 

Problem I. — To determine the instant of the Equinox, and 
the exact position of the Equinoctial points. 

Let the declination of the sun be observed daily with the 
mural circle, and the difference between the. right ascension 
of the sun, and some star, with the transit instrument, as the 
sun crosses the meridian, each day for some days, about the 
time of the equinox. 

Suppose M M 7 to 
to be the arc of the 
equator comprehended 
between the ideclina- 
tion circles M S and 
M 7 S 7 , passing through 
the sun at two noons 
near the equinox, the 
one before, the other 
after. The declination 
of the sun having 
$ ^ changed from north to 

south of the equator, or vice versa, in the interval,, as, from 
S to S', or from S 7 to S, M M' will be the difference of the 
right-ascensions of the sun, at the two observations ; each 
right-ascension of the sun having been estimated from- the 
declination circle passing through the star,. say a Lyrae as 
a zero. M S and M 7 S 7 will be the observed declinations of 
the sun, represented by £ and $': the corresponding right-as- 
censions, estimated from the equinoctial point E, in opposite 
directions, being a and a' . By the solution of the. right- 
angled spherical triangles, EMS and E M 7 S 7 , (Trigono- 
metry, p. 143.) 

__ sin a _ _ sin a' _ sin a tan d 

cot E= — — and cot E= : whence — — ,= ~, by 

tan tan o 1 sm a' tana 

equating the second members ; i. e., sin a : sin a' :: tan d •. 

tan <5 7 , or, by composition and division, sin a + sin a f : sin 

a - sin a :: tan d -j- tan d f : tan d - tan d r , or [Trigonometry, 




PROBLEMS. 9 

p. 78) tan d + tan $' : tan $ - tan <5' :: tan | (a + a 7 ) : tan 
^ («-a'). The last term of this proportion, being the only 
one unknown, may be computed : then | (a -f a') and \ 
(a - a') being known, by taking their sum and difference, a 
and a' will be obtained. The difference of right-ascension 
between the sun and equinoctial point, at either of the two 
noons, being now known, and the difference of right-ascen- 
sion between the sun, at this noon, and a Lyra?, being also 
known by observation, the difference of right-ascension be- 
tween a Lyra? and the equinox, becomes known. Such ob- 
servations should be repeated for several days before and 
after the equinox, and the mean of all the results be taken 
as the true difference of right-ascension between a Lyras and 
the equinox ; in other words, the true right-ascension of a 
Lyra?, estimated from the equinox as the zero. If now the 
clock be set to indicate the hours, minutes, and seconds, 
expressed by the right-ascension of a Lyra? in time, found, 
as before, at the instant that the star passes the meridian, 
the clock will indicate no hours, no minutes, no seconds (or 
the zero of time), when the vernal equinox comes to the 
meridian, and will indicate the right-ascension of any other 
heavenly body, in time, when that body comes to the 
meridian. 

Problem II. — Precession of the Equinoxes. 

If the autumnal equinox be observed in the same way 
as the vernal, its place will be found to differ 12 hours in 
right ascension, or exactly 180° from the vernal equinox; 
2. e., they are diametrically opposite to each other on a line 
passing through the centre of the earth, and extending to 
the celestial sphere. It will be found, in computing the dif- 
ference of right-ascension between the equinox and some 
star, like a Lyra?, as above, that the equinoctial point is not 
fixed, but that its difference of right-ascension from any star 
is all the while increasing* at a rate, if we take a mean of 
many years, of about 50. /r 2 in arc annually ; this is called 
" the precession of the equinoxes." The late Mr. Bessel, of 
Konigsberg, gives the exact expression for the mean preces- 
sion of one year, by the formula 50."21129-f-0.0002443£; 



10 ASTRONOMY. 

in which t denotes the number of years elapsed, at the time 
required, since 1750, A. D. 

Problem III. — The Calendar. 

The mean of a great number of intervals between two 
successive equinoxes, in time, is 365.242217 days, or 365d. 
5h. 48m. 58s. This is called a tropical year, and its exact 
length, as thus determined, is of importance in the regula- 
tion of the calendar, by which the labors of the husbandman 
are governed. If the year be supposed to consist of exactly 
365 days, as was the case before the Julian Calendar, then 
at the end of 4 years, the fraction by which the true tropi- 
cal year exceeds 365 days, amounts to nearly a day, wmich 
the sun has to move in declination, before its declination 
attains the same value it had attained on the same day of 
the month 4 years before. The improved Calendar, made 
by the decree of Julius Caesar, required the intercalation of 
one day in 4 years, by making 29, instead of 28 days in 
February, so that on the first of March, the sun would have 
attained the same declination that it had on the first of 
March 4 years anterior. But the Julian Calendar itself re- 
quired a correction in the lapse of centuries, owing to the 
error in taking the fraction over 365 days, in the length of 
the true tropical year, to be exactly 6 hours. This error 
was, however, the other way: 365 days were too little; 
365d. 6h. too much : the fraction by which it was too much 
accumulated so as to amount to 3 days in 400 years. In- 
stead, therefore, of intercalating a day at the end of Febru- 
ary, we omit to intercalate once in 100 years for 3 succes- 
sive centuries, as, for instance, in the years 1700, 1800, 1900 ; 
but in the fourth century, as 2000, intercalate, so as to omit 
three intercalations in 400 years, it requires 3500 years for 
the error of this mode of correction to amount to a day ; the 
correction being, truly, a little over 3 days in 400 years. 
This last reformation in the Calendar, is called the Gregorian, 
from Pope Gregory XIII., who made it in 1582. It is im- 
material, for the interests of agriculture, by what name any 
day in the year is called : it is only important that the day 
on which the sun attains a certain declination, north or 



PROBLEMS. 11 

south, should be called by the same name every year. Gre- 
gory chose to go back to the Council of Nice, A. D. 325, in 
which year the equinox happened on the 21st of March, and 
so to arrange as to have it happen on the 21st of March 
ever after. As the error in the interval from 325 to 1582, 
some 1300 years, at the rate of 3 days in every 400 years, 
amounted to 10 days, Gregory ordered the 5th of October, 
1582, to be called the 15th. This would make the vernal 
equinox happen on the 21st of March, 1583, as it did in 
325 ; otherwise, it would have happened on the 11th of 
March, 1583. This omission of 10 days all at once, made 
the difference between what is called the old and new style. 
The new style was not adopted at once in Protestant coun- 
tries ; in England, not till 1750, when 11 days were taken 
out, the 3d of September being called the 14th, owing to 
the intercalary day to be omitted in the year 1700. There 
is now a difference of 12 days between old and new styles, 
owing to the intercalary day that was omitted in 1800. 



Problem IV. — Determination of the Obliquity of the 
Ecliptic, 

The position of the equinoxes being accurately deter- 
mined by a method already given, a single observation of 
the right ascension and declination of the sun, will serve to 
determine the obliquity of the ecliptic (Trigonometry, p. 
143) ; the mean of a great many results should be taken. 
If the obliquity be computed from observations taken at 
very long intervals, it will be found to have undergone a 
change, of which physical astronomy teaches us the cause ; 
it results from the disturbing action of the other planets of 
the solar system. The obliquity is now diminishing ; after 
a very long period, it will begin to increase again ; and so 
will it continue to move back and forth for ever ; in the 
language of Pontecouland, " pendule immense qui batte des 
siecles "- — " an immense pendulum whose oscillations are 
GQtt£ii*uxms." The annual diminution of the obliquity, in 
the present age, is OZ'457. 




12 ASTRONOMY. 

Problem V. — To prove that the apparent orbit of the sun 
around the earth is equal, in all respects, to the real orbit 
of the earth around the sun. 

, •£ Let E E 7 be a portion 
of the real orbit of the 
earth, whatever be the 
curve, around the sun at 
S. When the earth is at 
E the sun is seen in the 
/jBJ direction E S. Suppose, 
now, the earth to have 
moved to E 7 , the sun will 
be seen in the direction 
E 7 S, the same direction 
as E S 7 parallel to E 7 S, i. e., the same direction as if the 
earth had continued stationary at E, and the sun had moved 
from S to S 7 . The two sectors, E S E 7 and S E S 7 , having 
their angles and radii equal, are equal ; their angles being 
angles which parallel lines make with E S, and their radii 
being fhe distance between the earth and sun. Since the 
same may be proved in the same way of all the elemental 
sectors of which the terrestrial and apparent solar orbits are 
respectively composed, it follows that the whole orbit of the 
earth around the sun, which is made up of the former, is 
equal to the whole apparent orbit of the sun around the 
earth, which is made up of the latter. 

The determination of the size and shape of the one 
orbit, then, will be the determination of the size and shape 
of the other. 

Problem VI. — The first approximation of the apparent 
solar orbit may be obtained as follows : 

Let the right ascensions and declinations of the sun be 
observed with the transit instrument and mural circle, and 
let the apparent diameter of the sun at the same time be 
observed with the micrometer, every few days throughout 
the year. Let the observed right ascensions and declina- 
tions be converted into longitude by the method pointed 



PROBLEMS. 13 

out ill Trigonometry, p. 143. Draw lines radiating from a 
point and making, with each other, angles equal to the dif- 
ference of longitude, found as above, throughout the year, 
and lay off on these lines distances proportional to the ob- 
served diameter of the sun at the same time (which ob- 
served apparent diameters will be inversely proportional to 
the distances of the sun from the earth) ; a curve traced 
through the points thus determined on these lines, will give 
the approximate form of the orbit, which will be found 
to be nearly an ellipse. 

Problem VII. — Continuation of former. 

Instead of this graphic construction, the radii-vectores of 
an ellipse, the pole being at the focus, with the semi-major 
axis unity, and the eccentricity equal to the ratio of the 
sum and difference of the greatest and least apparent diam- 
eters of the sun, may be computed for variable angles, equal 
to the difference of the sun's longitude when his semi-diam- 
eter is greatest ; and the various other values of the longi- 

«(1— e 2 ) 

tude by the formula r=— or, rather, since a is 

J \-\-e cos v, ' ' 

1— e 2 

unity, r-— ; and the length of these radii-vectores, 

J 1 -\-e cos v ' ° ' 

thus computed, will be found to bear the same proportion 
to unity that the corresponding apparent diameters do to 
half the sum of the greatest and least apparent diameters, 
thus identifying the orbit with the ellipse. If the areas of 
the sectors, comprehended between the radii-vectores, be 
computed, they will be found proportional always to the 
time employed by the radius-vector in describing them. 

The above investigation into the nature of the earth's 
orbit has developed two of the three celebrated laws of 
Kepler, which apply to the orbits of all the planets. 

Problem VIII. — Kepler^s Laws. 

1. The orbits are all ellipses, of which the sun is one of 
the foci. 



14 ASTRONOMY. 

2. The areas described by the radius-vector of a planet 
around the sun are proportioned to the times employed in 
describing them. 

3. The squares of the periodic times of the planets are 
as. the cubes of their mean distances from the sun, or the 
cubes of the semi-major axes of their orbits. That the semi- 
major axis is the mean of all the distances from the different 
points of the periphery of an ellipse to the focus appears, 
not only from its being half the sum of the greatest and 
least distances, but also from a well-known property of the 
ellipse, that the sum of the distances of any point from the 
two foci is constant and equal to the major axis. Owing 
to the symmetrical form of the ellipse, there will always be 
two points at the same distance from one focus that one of 
them is from the two foci. The points of the curve may r 
therefore, be arranged in pairs, the sum of the radii-vectores- 
in every pair being equal to the major axis, and the mean 
of the whole, therefore, will be equal to half this line. 

Problem IX. — Relation between the Radius Vector and the 

Velocity. 

As the areas described in equal times by the radius vec- 
tor, are equal (see Kepler's Law), the angular velocity of 
the radius vector at the perihelion must be greater, in order 
that the width of the sector described at that part of the 
ellipse in a given time, may be sufficiently greater to make 
up for its diminished length. There is, in fact, an exact re- 
lation between the angular velocity and the length of the 
radius-vector at all points of the orbit ; the former being 
inversely proportional to the square of the latter ; for, if v 
denote the angle described by the radius vector in a unit of 
time (as the distance passed over in a unit of time, is the 
definition of velocity, so the angle described in a unit of 
time, is the angular velocity), or the arc of a circle whose 
radius is unity which measures that angle, then r v would 
express the similar arc whose radius is r, since similar arcs, 
or arcs corresponding to equal angles in different circles, are 
to each other as their radii. If this r be the radius vector, 



PROBLEMS. 15 

and the small arc of the elliptical orbit described in a unit 
of time, say one day, be considered the arc of a circle, — 
which it may be, it is so small — the area of a corresponding- 
sector will be obtained by multiplying this arc r v by \r, 
so that we have the area of a sector described in a unit of 
time equal \r 2 v. Denoting this area described in a unit of 
time, by a, which, according to the second law of Kepler, is 
the same or constant in every part of the ellipse, we have 

2a 
a=h ,2 v, whence v = — ; i. e., the velocity varies inversely 
r 2 

as the square of the radius vector, the numerator in the 
value of v, being constant ; or, clearing the last equation of 
fractions, we have vr 2 =2<x, and if v and v" denote two par- 
ticular values of v, and r' and r" the corresponding values of 

I II o 

V T 

v" 2 whence — r.= -rrorV : v' r :: r" 2 : r' 2 
v r 2 , 



Problem X. — Longitude of the perihelion of the earth's or- 
bit, or longitude of the perigee of the solar orbit. 

The orbit of the earth, or apparent solar orbit, being 
symmetrical to the line of apsides, or major axis, the two 
registered longitudes of the sun which differ 180°, and the 
times of observations for which differ by half the whole 
^periodic time, will be the longitudes of the perihelion and 
aphelion points, or apsides of the orbit. No other straight 
Jine, than the line of apsides, drawn through the focus, 
divides the area of the ellipse into two equal parts. 

Note. — If the longitude of the perihelion and aphelion 
be determined, as above, at distant epochs, it will be found 
to have undergone a change in the interval, greater than 
would have been occasioned by the precession of the equi- 
noxes. The annual increase of the longitude of the perigee, 
or point of the apparent solar orbit nearest the earth, is at 
a mean 61. "9, and, as the retrograde motion of the equi- 
noxes is 50." 2 along these elliptic from east to west, there re- 
mains 11. "1 of actual progression in space, made by the 
perigee from west to east, annually. The anomalistic year 
is the time occupied by the sun in passing from the perigee 



16 ASTRONOMY. 

to the perigee again, or the interval between two consecu- 
tive times when the sun has the same anomaly. It exceeds 
the sidereal year, by the time which the sun would occupy 
in moving the llZ'7 above mentioned, in longitude ; and it 
exceeds the tropical year by the time occupied by the sun 
in moving 61/'9. The determination of the mean anoma- 
listic year, does not require a very accurate determination 
of the epoch, or longitude of the perigee. By taking the 
epoch of the perigee at very distant intervals of time, and 
dividing the whole interval expressed in days and fractions 
of a day, by the number of years intervening between the 
two epochs, the quotient will be the length of the mean an- 
omalistic year in days and fractions of a day. 

Problem XI. — Kepler^s Problem to determine the true 
anomaly from the mean by means of the eccentric. 

In the diagram, A M B is 
a semi-circle on the major axis 
A B as a diameter, the earth 
being at the focus E, and the 
sun at S in the apparent solar 
^B orbit ; the angle B E S is the 
true, and B C M the eccentric 
anomaly. 

In this problem we are to find the relation between the 
mean and eccentric anomaly. 

If the semi-axes of the orbit be denoted by a and 6, 
since the ordinate of a circle : the ordinate of the ellipse :: a 
: b (see Analytical Geometry), and since all the ordinates of 
the ellipse constitute the area of the circle, and since, by the 
theory of proportion, the sum of the antecedents : the sum of 
the consequents :: any one antecedent : its consequent, we 
have the area of the circle : the area of the ellipse :: a : b, and, 
for a similar reason, the area of the segment BMP: the 
area of the segment B8P::«:J,a similar ratio. The 
triangles E P M and EPS have also the same ratio, of a to 
6, since they have the same base E P, and are, therefore, to 
each other as their altitudes. The sum of the triangles and 
segments, or the sectors B E M and B E S, have, therefore, 

2ay<J^ / Cut T&* ^df^xZ^/ tf 
!%*, <&hC4s M wits 




PROBLEMS. 17 

the same ratio. If T denote the periodic time, or time of 
describing the whole ellipse, and t that of describing the 
sector B E S, and o=l, let the area of the circle be equal 
to <k ; and, since the areas are proportional to the times^the 
area of the sector B E M will be expressed by the fraction 

<k -. The area of the triangle MCE (if e denote the base 

C E, and u denote the eccentric anomaly, i. e., the angle C, 
or arc M B) is \e sin u. Adding the expressions for the 
areas of the sector and triangle, we have the area of the sec- 
tor B C M (which is also expressed, since a=l, by \ u) 

equal to *k ™~\-\ e sm u—±u, whence 2 *it -—u-esmii; but 

2 - expresses the angular motion of S about E, in a unit of 

time, say one day; i. e., the mean daily motion, if T be ex- 
pressed in days, because 2 ^ is the circumference whose 
radius is unity, and T the number of days in a complete re- 

2tf 
volution. Denoting — by n, nt will evidently be the mean 

anomaly, or mean angular motion, in the time or number of 
days expressed by t : the last equation thus becomes nt=u- 
e sin u (1.) or the mean anomaly in terms of the eccentric, 
nt being the mean anomaly, and u the eccentric. 

Problem XII. — To find the relation between the eccentric 
anomaly and the true anomaly. 

We have (by analytical geometry) the expression for 
the distance between Uyo points in space. 

{x—x"Yf{y—y"Y ; X 
If the two points be E and S, (see last diagram,) denoting 
the distance by r, namely, ES: x of the above equation in 
this particular case takes the value of e, y'=0', #"=cos u; 
b . 



y=-sine u=(Vl — e' 2 ) sin u; and ■& becomes 
a 

r 2 =(e — cos u) 2 ±(l — e 2 ) sin 2 ?/= 

e 2 — 2e cos u-\- cos 2 u-\- sin 2 u — e 2 sin 2 u. 
~ 2* 



18 ASTRONOMY. 

But sin 2 ^+cos 2 w = l ; therefore 

? ,2 =l-f-e 2 — 2e cos u — e 2 sine 2 u, or 
1+e 2 — 2eeosu — e 2 (l — cos 2 u) = l — 2e cos M-p-cos 2 ue 2 . 

Therefore r=l — 6 cos u. 
The polar equation of the ellipse, the origin of polar co- 

ordmates being at the focus, is in which v is 

l-\-e cos v, 

the true anomaly. Equating this value of r with that in 
the last equation, we have 

1 e 2 i e 2 

1 — e cos u= therefore \-\-e cos v= or, 

1+^cosv 1 — ecosw 

1 — e 2 - 1 — e 2 — l-f-ecosw eco&u — e 2 
e cos v= — 1= s ! = 

1 — ecosu 1 — ecos^ 1 — ecosw. 

cos u — e 



Dividing both sides by e, we have cos v== 



1 — ecos^. 



By Trig., (page 100,) tan 2 \ v= =,bvlast equation, 

1 -f- cos v 



- cos u- 



- e cos u 1 — e cos u — cos u -\-e 



1 _L cos ^ e 1 e COS U -f-COS U 



1 — ecosw 

•A 1 -X-p 

tan -2 u. Extracting square 



(1+e) (1 — cosw) 1+e a 



(1 — e)(l-[-costt) ~ 1 — e 



Vl+e 
X tan £ ^( (2), which 

expresses the relation between the true and the eccentric 
anomaly in each part. (In order to have convenient ex- 
pressions for computation, it is necessary to develop (1) and 
(2) in series, for which see Bowdich's "La Place," page 374, 
vol. i. Convenient practical methods are given in the 
"Theoria Motus Corporum Coelestium," of Gausse, pages 10 
and 13.) 



PROBLEMS. 19 



Problem XIII. 

What is termed the angle of eccentricity, is the angle 
whose sine is the eccentricity. If we denote this angle by 
d, we have by definition, e=sine ^ (1.) By Trig., page 104, 
(18,) we have Vl-e 2 =cos <p (2) or (y 1+e) (V l-e)=cos</> 
(3). From (1) we have l-e= l — sine (4). 

Dividing (4) by (3) ^_^ = 1 -^ =t an (46?-* *). 

(See Trig., page 102 (45). Then from (2,) last problem, 
tan -J- u 



tan (45°— $*). =tan £ v. 
Given ic— 324° 16' 29" 5, 0=14° 12' 1" 87. 

| ^=162° 8' 14'' 75 Log. Tan.= 9.5082198. 

45°— i?=Sl° 53' 59" 07 Log. Tan. =^8912 427. 

157° 30' 41" 5=Log. Tan. \ v= 9.6169771. 

Therefore v=315° V 23." 
To compute the corresponding length of the radius 
vector from the ellipse, which is the orbit. 

(1-e 2 ) 

The polar equation is r—a ) } ■ x If we denote the 

(1+e cos v) 

P 

numerator (called the parameter) hyp : then ?,ap J"V" 

Ji"T~ 6 COS t/, 

and then as 1 — e 2 =cos 2 </>,,by (2) above, therefore p=a 



cos 2 



To compute note. 
Given 0=14° 12' 1" 87, Log. Sine=Log. e=9.3897262 
^=315° V 25" Log. cosine=9.8496597 

e cos t--=0.l73545=Log. e cos ^=9.2393859 
Log. 1+e cos v=*Q. 0694959 
Given Log. a= 0.4224389 
Log. cosine 2 <£= 9.9730448 

Log.p= 0.3954837 

Log. 1 + e cos v= 0.0694959 



Log. r= 0.3259878 



20 ASTRONOMY. 



Problem XIV. — Gausse's Method of determining the eccen- 
tric anomaly from the mean. 

From (1) Prob. XL we have u=nt-\-e sin u. This equa- 
tion being transcendental, does not admit of direct solution; 
it may be resolved by tentative methods, assuming an ap- 
proximate value for u, and repeatedly correcting it until the 
equation shall be exactly satisfied. This, in practice, is much 
more convenient and expeditious than by a series as given 
by La Place. 

Let us suppose £ to be the approximate value of u and % 
expressed in seconds to be the value of the correction to be 
added, in order to have the true value of u. So then u—b 
-\-%. Let now the quantity e sin £ be computed in seconds, 
by logarithms, noting the difference of the logs, for one 
second in that part of the tables where we have sin e, and 
the difference of logs, corresponding to a difference of unity 
in the numbers in that part of the table, where we have the 
number e sin e, and its log. Let these differences be denoted 
respectively by I and ^. If now we suppose the value of £ 
to be assumed so near the value of u, that we may suppose 
the difference of the logs to be proportional to the difference 
of the numbers, it is evident that we may place e sin 

(e^-x}=e sin, £ ± — (the sign being . in first and fourth 

quadrant, and -& in 2d and 3d, because in 1st and 4th the 
sine is an increasing function of the arc.) Therefore, 
Since e-\-%=nt-{-e sin (£+%) or by last equation 
I % 
(0) £-\-x—nt-\-e sin £ ± — by transposing 

[X 

l X 

X^. — =nt-\-e sin £ — e, whence 
fi 

%= (nt + e sin £ — s) therefore by (0) above 

1 , 

i-\-%, or u=nt-\-e sin e ± -~y (nt-\-e sin £ — £.) 



PROBLEMS. 21 

Problem XV — Example, to obtain the eccentric from the 
mean anomaly. 

Given, the mean anomaly=332° 28' 54.77". 

%=14° 12' 1" 87. Log. Sine=9.3897262 
Radius of circle=206264"806. Log.=5.3144251 

- To6c>b" 

Eccentricity=e=14° 3' 20". Log. Sine=4.7041513 

For a first calculation, let us assume £=326°, as the ap- 
proximate value of the eccentric anomaly, which must be 
evidently less than the mean^360°— 326°=34°, and the 
log. sine of this is the same as the log. sine of fc£ 
Log. Sin e=9.74756 negative. 
Log. Sin 34° 1=9.74775 

Difference of logs. 1'= 19 

Dividing by 60 we get the difference of logs, for l 7/ = 
0.32 as the value of ^. 

Log. Sin £=9.74756 negative. 
Log. of e, in seconds, 4.70415 

Log. e sin e, 4.45171 negative. 
Thence e sin £=—28295' =7° 51' 35". 

Resuming the above logarithms, to get their differences, 
we have 

Log. 28295"=4.45l7l 
Log. 28285"=4.45155 



Difference, 10"= 16. Therefore, Indiffer- 

ence for one second, that is, ={i. The formula ^ — A=1.28", 

By hypothesis 72i=332° 28' 54.77" 

e sin £=—7° 51' 35''. Therefore, 



nt+e sin £=324° 37' 20". _ 
rV^1^#j?iC&£.= V 1C> 22' 40" =4960 seconds. 

Hence, — — x 4960 = 1240 seconds=20' 40" ; wherefore 
1.28" a 



22 ASTRONOMY. 

the corrected value of 76 is 324° 37' 20'' — 20' 40' =324° 
16' 40". 

To arrive at a greater accuracy, we repeat the same 
process with this result as a new assumed value, using the 
time tables of logs, to seven figures : 

Log. 6=4.7041513 
Log. Sin £ = 9.7663058 

n 



e Sin *, Log. =4.4704571=— 29543" 18=— 8° 23" 
18. Therefore, nt+e sin e = 324° 16' 31" 59. 

This differs from e by 8" 41, and this, multiplied bv 
I 29.25 

~ j=— z-yr, produces 2" 09 ; whence, finally, the cor- 
rected value of w, or the eccentric anomaly required, is 
324° 16' 31" 59— 2" 09 = 324° 16' 29" 5, exact within T fc 
of a second. 



Problem XVI. — Tables of the Sun. 

Tables of the sun have been computed by astronomers, 
which give his mean longitude at the beginning of a certain 
year, as 1800, and at the beginning of the preceding and 
following years. There are supplementary tables, by means 
of which the use of the tables may be extended to 10,000 
years before and after 1800. Tables of the sun also give 
the mean longitude of the perigee in the same way. In 
order to have the mean longitude at any instant, it is neces- 
sary to add to his mean longitude, at the beginning of the 
year, the mean daily motion in longitude, multiplied by the 
number of days and fractions of a day elapsed since the 
beginning of the year. This computation is facilitated by 
tables which give the motion of longitude from the begin- 
ning of each year to the first of each month ; then tables 
for days, which give the sun's mean motion in longitude for 
any number of days, from one up to thirty-one ; then tables 
for hours, giving the sun's mean motion in longitude, from 
one up to twenty-four ; then for minutes, from one to sixty ; 
and so for seconds. 



PROBLEMS. 2& 

Example. 

To have the mean longitude of the sun, June 20th, Oh. 
40m. 50s., 1853. 

Take out from the tables of years the mean longitude 
for 1853 ; also, from the tables of months, the mean motion 
in longitude against June ; from the tables of days, the 
mean motion in longitude against 20th ; and so on through 
the hours, minutes, and seconds. Add up all these num- 
bers, and the result will be the mean longitude of the sun 
at the instant required. The difference between the mean 
longitude of the sun and that of the perigee, at any time, 
is the mean anomaly at that instant. The tables of the 
sun give the equation of the center, corresponding to any 
given mean anomaly. (The tables are computed from 
formulae such as we have had already.) The equation of 
the center, applied as a correction to the mean longitude, 
gives the true longitude of the sun for a perfectly elliptical 
orbit (the orbit would be a perfect ellipse if there were but 
the sun and earth), estimated from the mean equinox. 

Note. — To have the true longitude from the true equinox, 
the equation of the equinoxes must be applied, of which here- 
after. 

Tables have also been computed for furnishing the effects 
of the perturbations by the different planets, in order that, 
being applied to the elliptical place of the sun, his true lon- 
gitude for the given instant may be obtained. These last 
are tables of double entry, as the disturbing effects of a 
planet will depend upon the relative position of the three 
bodies — the earth, sun, and planet ; the two arguments of 
the tables of perturbation are ephemeral. Finally, the 
value of the radius vector of the earth's orbit, computed on 
the supposition that the earth's orbit is a perfect ellipse, the 
semi-major axis of which is unity, is furnished by a table ; 
and other tables give the variations of that element, pro- 
duced by planetary perturbations. The value of radius 
vector for these tables may be computed from the polar 
equation of the ellipse, or from the equation r=a (1 — e 
cos w), u being the eccentric anomaly. 



24 ASTKONOMY. 



Problem XVII. — Orbit of the Moon. Longitude of the 
node and inclination. 

From the observed right ascension and declination of 
the moon, the longitude and latitude may be computed. 
(Vid. Trigonometry, page 139.) If, from a series of such 
observations, a record of corresponding latitudes be made, 
and if, on inspection, the latitude at any time be found equal 
to zero, the moon is then in the plane of the ecliptic, or in 
one of the nodes of her orbit. The corresponding longitude 
of the moon, given by the record, is the longitude of the 
node. The maximum of all the recorded latitudes would 
evidently be the inclination of the orbit. As no observation 
may give a latitude exactly zero, the instant of its being 
zero may be found by proportion, by means of two latitudes, 
the one a little north, the other a little south of the ecliptic. 
The proportion would run as follows : — As the whole 
change of latitude in the interval between two observations 
(equal to the sum of the north and south latitude) is to the 
interval of time between the two, so is the north latitude to 
the interval of time between the instant of observation fur- 
nished by the north latitude, and the instant that the moon 
is in the node. Then another proportion : — As the whole 
interval of time between the observations is to the whole 
change of longitude in the interval, so is the interval found 
by the preceding proportion, to the difference of longitude 
of the moon at the first observation and the node. The same 
method will apply to a planet. 



PROBLKMS. 



38 



Problem XVIII. — Method of determining the inclination and 
longitude of the node by two observations of the moon. 




Let E be the centre of the earth and the origin of three 
co-ordinate axes, the plaue of X Y being the plane of the 
ecliptic, and the axis of X the line of equinoxes, M the 
moon, E M the radius vector represented by r, E p projec- 
tion of the radius vector represented by r', the angle M E 
p or b will be the latitude, X E p or * tier longitude. We 
have evidently the following equations, since M p—z pf=y 
E 4 /==$, the co-ordinates of the moon, viz. : 



z =r sin. 



b 

?/ = /''sin. / 
or=?* / cos. I 

T =7' COS. b 



] Substituting in the second and third, 
{/'. its value obtained in the fourth. 
( x-==r cos b cos I 



for 



J y=> 



c< 
sin 



Q 



cos 
y=r cos b 
z=r sin b 

Three co-ordinates of the moon in space are thus ex- 
pressed in terms of her geocentric latitude, longitude, and 
radius vector. The general equate* of a plane passing 
through the origin of co-ordinates is A x + B y -f- C z = O. 

P O 
Dividing throughout by A, it becomes x + ry+ T" 2;= =0(a>) 



26 ASTRONOMY. 

Substituting for x, y, and z in this, their values given by 
equation (Q) for two places of the moon, and dividing 
throughout by r, we have two equations and but two un- 

R O 

known quantities, - and -. Substituting in (w) the values of 

A A 

R O 

- and - found from these two equations, we shall have the 
A A 

equation of the plane passing through the two positions of 

the moon and the center of the earth ; that is, the plane of 

the moon's orbit. Making z = in equation (o>), and re- 

solving it with respect to y, we have y =- x for the equa- 
tion of the trace on the plane of X Y. This trace, which is 
the line of nodes, makes an angle whose tangent is 

— with the axis of X or line of equinoxes. The value of A 

and B having been already found, as shown above, the 
angle wdiieh the line of nodes makes with the line of equi- 
noxes ; in other words, the geocentric longitude of the node 
becomes known. As for the angle w 7 hich a plane makes 
with the plane of X Y, we have (vid. Analytical Geometry) 

cos. angle = — , and this is the inclination of the 

Vl+B 2 +C 2 
plane of the orbit with the plane of the ecliptic. 

Problem XIX. — Retrogradation of the Moorfs node. 

The longitude of the node will be found (as above) to be 
different at every revolution of the moon. The node moves 
back from east to west, and this is called the retrogradation 
of the moon's node. This retrogradation is found to amount 
to 360°, or, the line of nodes makes a complete revolution 
about the earth, so that the longitude of the node becomes 
the same again, estimated from the true equinox in 6798 
days, 12 hours, 57 minutes, 52 seconds, or in about 18f 
years. The physical cause of this phenomenon is the dis- 
turbing action of the sun. The sun's action, when the moon 
is out of the plane of the ecliptic, may be decomposed into 



PROBLEMS. 27 

two, one acting in the plane of the ecliptic, the other perpen- 
dicular to it, which latter component has a tendency to 
draw the moon to this plane sooner than it would reach it 
if it had been undisturbed, and this occasions a precession or 
retrogradation of the nodes. The precession of the equi- 
noxes may be explained in the same way, if we suppose a 
number of moons following each other in the same orbit, 
each of these will follow the above law, and this would be 
the case if the moons were contiguous, forming a ring. 
The same phenomenon would occur, though in a less degree, 
if the ring were diminished till it clasped the earth. Now, 
the earth is a spheroid, having its polar diameter shorter 
than its equatorial : it may, therefore, be regarded as a 
sphere clasped by a ring of matter, the thickness of which is 
greater at the equator, and diminishes to at the poles. 

Problem XX. — Nutation. 

The action upon the protuberant mass at the equator, 
producing the phenomenon of precession, is not the action 
of one body alone, but the combined action of the sun and 
moon, and must depend on the relative position of the sun, 
moon, and earth. This relative position must become the 
same after each revolution of the moon's nodes ; there must, 
therefore, be a variation of the mean amount of precession, 
which goes through a period of 18f years. This variation 
is called nutation, and the correction to be applied in con- 
sequence of it, given by the tables, is called the equation of 
equinoxes. The phenomenon of nutation was first discov- 
ered by Dr. Bradley. He found that the star Gamma Dra- 
conis varied its apparent place, or right ascension and de- 
clination, and that this variation completed its period in the 
time of revolution of the moon's node, or in 18f years ; and 
that the variation could be accounted for by supposing that 
the true pole of the heaven described a small ellipse about 
the mean pole, the semi-axes of which are 9" and 7". 

Problem XXI. — Aberration of hight. 

This phenomenon was also discovered by Dr. Bradley, 
at the same place and with the same instruments. A va- 



28 ASTRONOMY. 

nation in the right ascension and declination of a star, as 
Gamma Draconis, was noticed to complete its period in a 
year, and could be accounted for by supposing that the ap- 
parent place of the star described a small ellipse about its 
true place, every year, whose major- axis is 20", and whose 
minor-axis is to the major as the sine of the stars' latitude 
is to unity. A phenomenon thus evidently connected with 
the motion of the earth in its orbit, and which may exactly 
be accounted for by supposing it to arise from the com- 
bined motion of the earth with that of light, (the velocity 
of light, previously discovered by means of the eclipses of 
Jupiter's satellites, to be such as to pass over the diameter of 
the earth's orbit into 16" of time : by comparing the ob- 
servations of the eclipses of the satellites when Jupiter was 
in opposition, with those when he was in conjunction, i. e. 
when the earth was on that part of its orbit nearest to Ju- 
piter, and when on the opposite side of the orbit most re^ 
mote from Jupiter.) The effect of these combined motions 
may be illustrated by rain falling vertically, which, to a per- 
son in motion, appears to fall obliquely towards him ; so 
light, coming from a star to an observer in space with the 
earth in its orbit, will appear to come from a point a little 
forward of the true place of the star. A parallelogram of 
velocities being constructed, with its sides proportioned to 
the relative velocities of light and the earth, the diagonal of 
this parallelogram will make an angle, with its longest side, 
equal to the angular displacement of the star. As the earth is 
moving, in all possible directions in space, in the course of a 
year, the displacement of the star will be in all directions 
from its true place, and will, evidently, describe an oval 
curve about its true place in the heavens. If a star be in 
the pole of the ecliptic, its displacement, occasioned by the 
motion of the earth in the plane of the ecliptic, in a direc- 
tion nearly perpendicular to the light from the star, will be, 
throughout the year, the same ; and the star will appear to 
describe a circle about its true place equal to the maximum 
of aberration, or constant of aberration, = 20". If the star 
be in the plane of the ecliptic, the parallelogram of veloci- 
ties being, in this case, always in this plane, the angle of dis- 
placement will be always in this plane, and, consequently, 



PROBLEMS. 29 

the ellipse of aberration degenerates into a straight line. If 
the star be anywhere between the plane of the ecliptic and 
its pole, the ellipse will be more or less eccentric. (Nuta- 
tion or aberration, in right ascension and declination, or in 
latitude and longitude, i. e. the effect of these phenomena 
upon these co-ordinates, being very small, formulae for their 
computation may be best obtained by aid of the " Differen- 
tial Calculus." See Gauss' " Theoria Motus Corporum Cce- 
lestium.") 

Problem XXII. — Continuation of the determination of the 
elements of the Moon's orbit. 

By the solution of a right-angled spherical triangle, of 
which the inclination of the moon's orbit is one of the oblique 
angles, and the difference of longitude of the moon and the 
node of her orbit the base ; or instead of the latter using the 
latitude of the moon, as the other perpendicular side of the 
triangle, the hypothenuse may be computed, which will be 
the angular distance of the moon from the node. 

Note, — The retrogradation of the node need not be 
noticed in the computation, as it has no effect on the quan- 
tity obtained. If a series of such angular distance be com- 
puted from the observed right ascension and declination of 
the moon when on the meridian, and the interval of time 
between the observations exactly noted, the daily angular 
motion of the moon in her orbit may be obtained. The 
relative distances of the moon from the earth may be ob- 
tained approximately by means of her apparent diameter ; 
or the absolute distances from her horizontal parallax. 
By means of the moon's daily angular motion in her orbit, 
and her daily variations of distance from the earth, the 
ellipse, w r hich is her orbit and the position of perigee, may 
be obtained in a manner precisely analogous to that em- 
ployed for the orbit of the earth, or apparent orbit of the 
sun. 



3* 



ASTRONOMY. 



Problem XXIII. — Method of determining the horizontal 
parallax of the Moon. 




Suppose that two observers in distant latitudes, and, for 
the sake of simplicity, on the same meridian, observe - the 
difference of zenith distances, between the moon and some 
fixed star. Let A and B be the two places of observation, 
M the moon, A S and B S' parallels in direction of the star 
at an infinite distance. Then (Trigonometry, p. 27o) since 
A M C and BMC are the parallaxes in altitude, we have 
the horizontal parallax H=A M C -^ sin M A Z. Therefore 
H x sin M A Z=A M C. Similarly, at station B, H x sin 
M B Z'=B M C : by addition, H (M AZ + MBZ') = AMC 
4- B M C 

- 1 -tt A MC + BMC _ A .M B _ 

sin M A Z 4- sin M B Z' last denominator. 
AMB-S' + M B S'=M AS + MBS'; i.e. the sum of 
the differences of zenith distances of the moon and star. If 
the observers are not on the same meridian, allowance must 
be made for the change of the moon's declination in the 
interval of her transits over the two meridians (given by 
Nautical Almanac). The observed zenith distances should 
be reduced to geocentric zenith distances, by a method simi- 
lar to that for converting astronomic to geocentric latitude. 
(Trigonometry, page 365.) 



PROBLEMS. 



31 



Problem XXIV. — Computation of the elements of a, Lunar 

Eclipse. 



sM 



LetXMVbe a portion of the moon's orbit, so small 
that it may be regarded as a straight line, M the position of 
the moon at any instant, N near, say, the even hour before op- 
position, M q a portion of a parallel of declination, passing 
through the moon, S the place of the centre of section of 
shadow, at the distance of the moon, at the same time, S q 
a declination circle passing through it. The declinations 
and right ascensions of the moon and center of the shadow- 
are given at this instant by the Almanac, the latter differing 
from that of the sun by 180° in right ascension, and being 
the same with contrary signs in declination. In the right- 
angled triangle M s q, there are known S <?, the Diff. of Decl. 
of the moon and center of the shadow, M q, their DirT. of R 
A, multiplied by the cosine of the moon's Dec, with which 



32 ASTRONOMY. 

the other parts of the right-angled triangle may he com- 
puted. Let m n denote the moon's relative hourly motion 
in declination, i. e. the difference of hourly motion of the sun 
and moon, (given in Almanac,) M m = corresponding hourly 
motion of the moon on a parallel of declination M q, which 
is obtained by multiplying the difference of hourly motion 
of the sun and moon in right-ascension (given by Almanac) 
by cos. of moon's declination. 

By solution of the right-angled triangle M m n, which 
may be considered as a plane-triangle, the two perpendicu- 
lar sides of which are thus given : we get M n, the moon's 
relative hourty motion in her orbit, and the angle m M w, 
which, subtracted from the angle S M q, before found by 
the solution of the triangle S M #, gives us the angle S M 
V, and consequently its supplement S M X, and then in 
the triangle S M X, having two sides and an angle, namely, 
S M, already computed, from triangle S M q, the side S X 
= the sum of the semi-diameters of the moon and section 
of shadow (the former of which is given by the Almanac, 
and the latter depends upon the parallaxes of the sun and 
moon, also given by the Almanac) ; we may compute M X ; 
the line, M X -4- M n will give the hour and fractions to be 
applied to the instant the moon was at M, in order to have 
the instant of first contact of moon and shadow — completing 
the isosceles triangle S X Y, the center of the moon will 
evidently be at Y at the instant of last contact ; then, 

MY. 

— — will be the hours and fractions of an hour from the 
M n 

instant the moon is at M, to the instant of last contact, and 
M V -*- M n will be the hours and fractions from the in- 
stant the moon is at M, to the instant of greatest phase. 



PROBLEMS. 



33 



Problem XXV. — Determination of the semi-diameter of the 
section of the shadow of the Earth at the distance of the 
Moon. 



DIAGRAM. 




The interior angle C is equal to the exterior angle S, minus 
the other interior angle p, and the interior angle 2, equals 
the exterior angle P, minus the interior angle C. Substi- 
tuting the value of C above, we have 2 = P + p — S. 

2 sss semi-diameter of the shadow required. 

S = semi-diameter of the sun, i. e , the visual angle at 
the earth, subtended by the radius of the sun. 

p = horizontal parallax of the sun or angle at the sun, 
subtended by the semi-diameter of the earth. P,£> and$ 
are given by the Nautical Almanac. P = horizontal pa- 
rallax of the moon. 



34 



ASTRONOMY. 



Problem XXVI. — Determination of the longitude of the 
node and the inclination of a plane fs orbit. 



DIAGRAM. 

Let the upper ellipse be the 
planet's orbit, and the lower, the 
projection of the orbit on the plane 
of the ecliptic. S, the place of the 
sun ; S N, the line of nodes. The 
longitude of IS" being determined 
by observations, (t. e., by marking 
the time when the latitude is equal 
to zero,.) wait till the geocentric 
longitude of the sun is equal to 
the longitude of m the node; the 
earth will then be on the line of 
nodes. Suppose it to be at E; 
let P be the place of the planet 
at this time ; let fall P p perpen- 
dicular to the plane of the ecliptic ; 
draw p A perpendicular to E N, 
and join P" and A. The angle P A p is the inclination of 
the planet's orbit to the ecliptic (by definition). Then 
N E it is equal to the difference of longitude of the sun 
and the node of the planet, and the geocentric longitude of 
the planet ; P E p, is equal to the geocentric latitude of the 
planet. In the triangles VEp and E A p right-angled at 
M and A, we have F p = Ep tan P E p. Ap = E p 
/sine NE_p. 

Pp 
Therefore, dividing the one by the other- — = tan PA 

Ap 

tan P E p . _ _ ,,,.,. 

= -T- ^-^; — or an expression tor the value ot the lnclina- 
sin N E p r 

tion of the planet's orbit in terms of the latitude and 

longitude. Remark. — The Heliocentric longitude of the 

node is the same, or differs from that of the earth when 

on the line of nodes, by 180 degrees. 




PROBLEMS. 



35 



Problem XXVII. — Determination of the elements of a 
planetary orbit. 



/ 



r' 




Let p be the place of a planet, p' its projection on the 
plane of X Y, for which, take the plane of the ecliptic and 
form the axis of X, a line through the center of the earth 
parallel to the line of nodes. The line E °p being the line 
of equinoxes, also in the plane of X Y. Denote the angle 
of these two lines or the heliocentric longitude of the node 
by n, the geocentric longitude of the sun by L, the geocen- 
tric longitude of the planet by Z, its geocentric latitude by 
I, the distance of the earth from the sun or radius vector 
of the terrestrial orbit by R, the distance of the planet from 
the earth by r, its projection by r", from the sun or the 
radius vector of the planetary orbit, by r ; the co-ordinates 
of the sun, by X. Y, of the planet, by x. y. z. Then, by draw- 
ing perpendiculars from S and p to the axes of X and Y, 
we shall have right-angled plane triangles formed, which 
will give us the following : 



36 ASTRONOMY. 

X = Rcos (t-»),Y = Rsin (L-n) (1) 

x = r" cos (I - 11), y = r" sin (I - n) (2) 

And from the right-angled triangle p E p 

z = r sin ^, r" = r cos A (3) 

Substituting in Equation (2) for r" its value in the second of 
equations (3) and writing after the results the first of equa- 
tions (3) 

x = r cos A cos (l-n) ) 

y = r cos I sin (I - n) > (a) 

z = r sin ^ ) 

Transfer now the origin of co-ordinates from E to S, i. e. y 
from the center of the earth to the center of the sun, and 
let the new axes be parallel to the old. Denoting the co- 
ordinates of the planet referred to, the new origin by x y z\ 
the formulas for transformation (see Analytical Geometry) 
are, since X and Y are the co-ordinates of the new origin, 

x = X + x<,y=Y+y\z = z 
.". x — x - X, y = y — Y, z'= z. 
Substituting for X and Y their value, equation (1), and for 
x. y. z, their value in (a), these last become 

x = r cos I cos (I - n) — E cos (L-n) ) 

y = r cos I sin (I - n) — R sin (L - n) > (b) 

z 1 = r sin I ) 

Squaring each of the group (6), and adding, observing that 



x 



1 2 



~r y 12 +s 12 = r' 12 , we have r 12 , = r 2 cos 2 I [cos 2 
(Z - n) + sin 2 (f - n) ] + R 2 [cos 2 (L - n) + sin 2 (£- w) ] 
+ r 2 sin 2 I — 2 r Rcos I [cos (l~n) cos (L-n) -j- sin (£-ft) 
sin (L-w)]. But since sin 2 -f- cos 2 of any arc = 1, the 
multipliers of R 2 and r, equal to unity. The equation thus 
reduces to - 

r' 2 = r 2 (cos 2 A + sin 2 A>R 2 — 2 r R cos A [ (l- n) 
-(L-n)] 
or r' 2 = r VR 2 — 2 r R cos I cos (Z - L) (c) 

We may have another expression for the line pp' = z' . 
For if we conceive a perpendicular from p to the line of 
nodes which runs through S parallel to the axis of X, this 
perpendicular will be equal to a^H^ an d multiplied by the 
tang, of the inclination of the orbit, will give pp, in symbols, 
t denoting the inclination 

z' = ffe - 3£) tan i. 



PROBLEMS. 37 

Substituting the values of z\ #, X, given by (a) (b) (1), this 

becomes s^ ^^ 

r sin I = r cos ^ eas (I -n) — R ees (L - n) tan i. 

This equation and equation (c) containing only two unknown 

quantities r and r\ these values may be found, and r being 

known, x\ y and z' the co-ordinates of the planet become 

known. 

But x is the projection of r on the axis of X ; therefore, 

x 

- is the value of the cosine of the angle which / makes 

with this axis, i. e., the angle which the radius vector makes 

with the line of nodes, or the angular distance of the planet 

from the node. 

The line r" being known from the geocentric k£*gii»de, ; 

and r by the last of equations (3) or by the equation r" 2 = 

r 2 - z' 2 . Moreover, x is the projection of r on the axis, 

x' 
therefore —7 is the cosine of the angle which the projection 

of r makes with this axis, or the heliocentric longitude of 
the planet, estimated from the line of nodes. 



Problem XXVIII. — Determination of the semi-major axis, 
eccentricity and longitude of the perihelion of a planets 
orbit by means of three radii vector es and the corres- 
ponding angular distances of the planet from the node. 

Let r. r'. r" be the three radii vectores, 

v v 1 v" the angular distances from node. 

ne the angular distance of the perihelion from the 

node. 
a the semi-major axis. 
e the eccentricity. 
The polar equation of the ellipse (by Analytical Geo- 
metry) is 

1 \e cos v 
in which v denotes the true anomaly or angular distance of 



.38 ASTRONOMY. 

the planet from the perihelion, equal to the difference of the 
angular distances of the planet and perihelion from the 
node. Clearing eq. (1) of fractions, and substituting for v 
its value for each of the three observations, we have 

1 + ecos (*-*) = a (l-e 2 )- (2) 

1+ecos (*'-*) = a (l-e 2 )i (3) 

1+eeos (*"-*) = a (l-e 2 )-^ ( 4 ) 

Subtracting (2) from (3) and (2) from (4) 

e [cos (*' - «r) — cos (v - «) ] = a (1 - e 2 ) (V-J) (5) 

< [cos (*"-*) - cos {v - *)| = a (1 -*) (77-;) (6) 

Dividing (5) by (6), and writing Q for the quotient of the 
second members, which will be a known quantity, since r, 
r', r*' are supposed to be known, we have 

cos (v r - <ii) — cos (v - #) _ 

cos (v" -ir) — cos (v - nr) 
or by Trig,, p. 101, (26) 
sin ^ (v' f v - 2 #) sin J (^ - *') __ 
sin 1 (""+"- 2 at) sin £ (* - *") ~~ ^ 

Multiplying by the last factor of the denominator, and 
dividing by the last of the numerator, both known, the re- 
sult in the second member will be known, and we have 

sin \ jV -f v - 2 jr) __ 
sin J (* 7/ + * - 2 at) " ' 

Putting a for \ (»' + v) and a' for £ (*" -f- *) 
sin (o- - «7r) = R sin (a' - #) 

or sin a cos rf — cos a sin it = R (sin a' cos *x — cos <*' 
sin tf) dividing throughout by cos *tt. 

Sin o- — cos a tan ^ = R (sin v r — cos or 1 tan #). Tan * 
being the only unknown in this last equation, its value may 



PROBLEMS. 39 

be found, and being substituted in either two of eqs. (2) (3) 
(4), the values of a and e may be found from them. 

The angular distance of the perihelion from the node, 
being known, by the solution of a right-angled spherical 
triangle, of which this angular distance is the hypothenuse, 
and the inclination of the orbit one of the oblique angles, 
the side adjacent this angle may be computed, and will be 
the longitude of the perihelion estimated from the node, by 
adding to which, the longitude of the latter, that of the 
perihelion will become known. 



■\4*C\ 






LIBRARY OF CONGRESS 



003 538 550 2 



CHARLES C. SUYDAM, 

WILLIAM J. SEABURY, 

WILLIAM T. VAN RIPER, 

CHARLES N. CLARK. 



> * 



